tag:blogger.com,1999:blog-54828016115016737042023-06-20T05:28:37.862-07:00Calculus IITillyhttp://www.blogger.com/profile/16058647084628708329noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-5482801611501673704.post-24474602459406849782011-02-10T20:47:00.001-08:002011-02-10T20:47:17.214-08:00Busy day today :/ More will be posted tomorrow.Tillyhttp://www.blogger.com/profile/16058647084628708329noreply@blogger.com1tag:blogger.com,1999:blog-5482801611501673704.post-57693597295684686952011-02-09T20:28:00.000-08:002011-02-09T20:28:41.222-08:00FINALLYFinally, after hours of work, I have figured out the secret to trigonometric substitutions.<br />
<br />
The biggest tip I can give is to restrict the domain of your sinθ, secθ, and tanθ functions.<br />
<br />
For sinθ -π/2 <u><</u> θ <u><</u> π/2<br />
<br />
secθ 0 <u><</u> θ < π/2 OR π <u><</u> θ <u><</u> 3π/2<br />
<br />
tanθ -π/2 < θ < π/2<br />
<br />
<br />
NOTE: Be cautious when it comes to reference triangles. Most of these will only need one reference triangle due to the restriction of the domain.Tillyhttp://www.blogger.com/profile/16058647084628708329noreply@blogger.com5tag:blogger.com,1999:blog-5482801611501673704.post-30740384673099457442011-02-09T14:37:00.000-08:002011-02-09T14:37:37.205-08:00Before I start to go over what I've been doing in regards to trigonometric substitution, it'd be helpful to explain integration by parts. <br />
<br />
Integrating by parts is based off the product rule and can be summed up like this:<br />
<br />
where int = an integral.<br />
<br />
int( u dv) = uv - int( v du).<br />
<br />
Essentially you break down the initial integral into the parts u and dv with dv being the most complicated function that you are able to easily integrate.<br />
<br />
It's best to set it up like so:<br />
<br />
u = ???? dv = ????<br />
du = ???? v = ????<br />
<br />
If you maintain this order, it's easy to integrate regardless of what letters you are using for u and v.Tillyhttp://www.blogger.com/profile/16058647084628708329noreply@blogger.com1tag:blogger.com,1999:blog-5482801611501673704.post-25480641208510190622011-02-09T13:42:00.000-08:002011-02-09T13:42:05.249-08:00I'm having issues remembering the domains and ranges of the inverse trigonometric functions, so I think it would be helpful to anyone reading that I post them here.<br />
<br />
arccosx Domain: -1 <u><</u> x <u><</u> 1<br />
Range: 0 <u><</u> y <u><</u> π<br />
<br />
arcsinx Domain: -1 <u><</u> x <u><</u> 1<br />
Range: -π/2 <u><</u> y <u><</u> π/2<br />
<br />
arctanx Domain: all values of x<br />
Range: -π/2 < y < π/2<br />
<br />
arcsecx Domain: all values of x except -1 < x < 1<br />
Range: 0 <u><</u> y <u><</u> π arcsecx =/= π/2<br />
<br />
arccscx Domain: The same as arcsecx<br />
Range: -π/2 <u><</u> y <u><</u> π/2<br />
<br />
arccotx Domain: All values of x<br />
Range: -π/2 < y < π/2<br />
<br />
I've got copious amounts to do tonight, so I'll probably be posting random tidbits all evening.Tillyhttp://www.blogger.com/profile/16058647084628708329noreply@blogger.com0tag:blogger.com,1999:blog-5482801611501673704.post-32718829330501952292011-02-09T13:17:00.000-08:002011-02-09T13:17:31.795-08:00Calculus IIHello all! This is my blog pertaining to what I'm learning in Calc II currently and what I'm confused with and working on.<div><br />
</div><div>Currently, we are working on trigonometric substitution for integrals.</div><div><br />
</div><div>I will post more after I finish my homework! </div>Tillyhttp://www.blogger.com/profile/16058647084628708329noreply@blogger.com6